R-form can turn an equation containing both sine and cosine into an equation involving one trigonometric function. The method has two distinct stages: combine the expression correctly, then solve the resulting shifted trigonometric equation over the required interval.
1. Combine before solving
Consider 3sin x+4cos x=2. Write
3sin x+4cos x=5sin(x+α), where cosα=3/5 and sinα=4/5.
The equation becomes 5sin(x+α)=2, or sin(x+α)=2/5.
2. Transform the interval with the angle
If the original interval is 0≤x≤2π, then the shifted variable θ=x+α satisfies α≤θ≤2π+α.
Solve sinθ=2/5 over that transformed interval, then subtract α from every allowed θ-value.
This is safer than solving over 0 to 2π for θ and assuming the same solutions survive the phase shift.
3. Range test before inverse trigonometry
For a sin x+b cos x=c, compute R=√(a²+b²). If |c|>R, there is no real solution because the combined sinusoid cannot reach c.
Example: 5sin x+12cos x=14 has no real solution because R=13.
4. Worked exact equation
Solve sin x+cos x=1 for 0≤x≤2π.
sin x+cos x=√2sin(x+π/4). Therefore
√2sin(x+π/4)=1, so sin(x+π/4)=√2/2.
Let θ=x+π/4. Then π/4≤θ≤9π/4. Within this interval, sinθ=√2/2 at θ=π/4,3π/4,9π/4.
Subtract π/4: x=0, π/2, 2π.
Both endpoints are included, so x=0 and x=2π are both valid inputs.
5. Worked equation with a vertical constant
Solve 3sin x+4cos x+1=6.
First move the constant: 3sin x+4cos x=5. R-form gives 5sin(x+α)=5, so sin(x+α)=1.
Therefore x+α=π/2+2πn, hence x=π/2−α+2πn. A stated interval would determine which members to retain.
6. R-form can reveal tangency-like equality at an extreme
When |c|=R, the shifted sine or cosine must equal ±1. Over each full period there is one phase position for the maximum and one for the minimum.
When |c|<R, there are generally two phase positions per full sine or cosine cycle.
7. Degrees work the same way
If 3sin x+4cos x=5sin(x+α) and x is measured in degrees, solve the shifted equation using degree periodicity. The algebraic R-form does not depend on the angle unit, but the numerical phase and period notation do.
8. Negative coefficients need phase control
For 5sin x−12cos x, one convenient form is 13sin(x−β), with cosβ=5/13 and sinβ=12/13.
Writing 13sin(x+β) would produce a positive cosine coefficient and therefore solve a different equation.
9. Graphs provide an independent check
The equation a sin x+b cos x=c asks where a sinusoid of amplitude R intersects the horizontal line y=c. The number of expected intersections over a full period can be predicted from the range test.
This is useful for detecting a missing quadrant solution after inverse-trigonometric work.
10. Common mistakes
- Solving for x before finishing the R-form conversion.
- Using the wrong phase sign.
- Forgetting to shift the interval together with x.
- Reporting only the principal inverse-trig value.
- Attempting inverse trig when |c|>R and no real solution exists.
11. Practice
- Use R-form to decide whether 8sin x+6cos x=11 has a real solution.
- Solve sin x+cos x=0 for 0≤x<2π using R-form.
- Write 5sin x+12cos x=13 as a shifted-sine equation and state the condition on its shifted angle.
- Explain why 7sin x+24cos x=−26 has no real solution.
- For 3sin x+4cos x=2, state R and the transformed equation before solving.
12. Answers
- R=10, and |11|>10, so no real solution.
- √2sin(x+π/4)=0. Thus x+π/4=nπ. On 0≤x<2π, x=3π/4,7π/4.
- R=13. Write 13sin(x+α)=13 with cosα=5/13, sinα=12/13; hence sin(x+α)=1.
- R=25, so the expression cannot be less than −25.
- R=5; 5sin(x+α)=2, so sin(x+α)=2/5 with cosα=3/5 and sinα=4/5.
Method selection
R-form is especially useful when the equation contains a linear combination of sine and cosine with the same angle. If the equation is already a quadratic in sin x, cos x or tan x, algebraic substitution may be shorter. Choose the representation that reduces the number of moving parts.
Continue with Solving Trigonometric Equations Over a Given Interval, Deriving R-Form Step by Step, or return to the Additional Mathematics Hub.